10.2-7题：给出一个时间复杂度为O（n）的非递归过程，实现对一个含n个元素的单链表的逆转。
程序的主要思想就是，转变指标的指向，例如原本是1->2->3->4，现在变成1<-2<-3<-4，就实现了逆转
package main
import (
"fmt"
)
type Node struct {
next *Node
data int
}
type List struct {
first *Node
}
func (l *List) Insert(d int) {
node := &Node{nil, d}
node.next = l.first
l.first = node
}
func revise(l *List) {
if l.first == nil {
return
}
curr := l.first
next := curr.next
curr.next = nil
for {
last := curr
curr = next
next = curr.next
curr.next = last
if next == nil {
l.first = curr
break
}
}
}
func printList(l *List) {
if l.first == nil {
fmt.Println("the list is empty")
return
}
fmt.Print("list: ")
for node := l.first; node != nil; {
fmt.Printf("[%d]", node.data)
node = node.next
}
fmt.Print("\n")
}
func main() {
list := &List{nil}
for i := 0; i < 10; i++ {
list.Insert(i)
}
printList(list)
revise(list)
printList(list)
}
